The distance between the circumcentre and the ortho-centre of the triangle formed by the points (2, 1, 5), (3, 2, 3) and (4, 0, 4) is
A.\[\sqrt{6}\]
​
B. \[\dfrac{\sqrt{6}}{2}\]

C. \[2\sqrt{6}\]

D. 0

Hint: -In this question, firstly, we will find the length of the sides of the triangle that is formed by the points (2, 1, 5), (3, 2, 3) and (4, 0, 4) by using the section formula that is given as follows
The most important formula that would be required to find the distance between two points in 3-dimension is the distance formula which is as follows
The distance formula is as follows
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\]

Complete step-by-step answer:
As mentioned in the question, we have to find the distance between the ortho-centre and the circumcentre of the triangle formed by the points that are given in the question.
Now, as mentioned in the hint, we will first find the distance between the points using the distance formula as follows
Distance between the points (2, 1, 5) and (3, 2, 3) is as follows
\[\begin{align}
  & \left[ d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} \right] \\
 & \Rightarrow d=\sqrt{{{\left( 2-3 \right)}^{2}}+{{\left( 1-2 \right)}^{2}}+{{\left( 5-3 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{1+1+4} \\
 & \Rightarrow d=\sqrt{6} \\
\end{align}\]
 Distance between the points (2, 1, 5) and (4, 0, 4) is as follows
\[\begin{align}
  & \left[ d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} \right] \\
 & \Rightarrow d=\sqrt{{{\left( 2-4 \right)}^{2}}+{{\left( 1-0 \right)}^{2}}+{{\left( 5-4 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{4+1+1} \\
 & \Rightarrow d=\sqrt{6} \\
\end{align}\]
Distance between the points (3, 2, 3) and (4, 0, 4) is as follows
\[\begin{align}
  & \left[ d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} \right] \\
 & \Rightarrow d=\sqrt{{{\left( 3-4 \right)}^{2}}+{{\left( 2-0 \right)}^{2}}+{{\left( 3-4 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{1+4+1} \\
 & \Rightarrow d=\sqrt{6} \\
\end{align}\]
Now, as the distance between all the points of the triangle is equal, hence, the triangle is an equilateral triangle.
Now, we know that in an equilateral triangle, all the points namely orthocenter, circumcentre and centroid lie at the same point, hence, the distance between the ortho-centre and the circumcentre (which is asked in the question) is 0.
Hence, the answer to the question is 0.

Note:- It is very important for the students to know that in an equilateral triangle, all the points namely orthocenter, circumcentre and centroid lie at the same point as if the students don’t know this, then they won’t be able to get to the correct answer.
Also, the students must know the formula that is used to find the distance between two points in 3-dimension which is the distance formula as without knowing them, one can never get to the correct answer.